This post includes a third proof that the absolute Russell set exists. The third proof invokes an idea that was used in the second proof, but is simpler and more direct than the second proof. The idea is expressed in the following quotation from the argument for Postulate 2 in the second proof; the second proof is located at viewtopic.php?p=2699904#p2699904.

browser32 wrote:A conditional statement with a necessarily false hypothesis is only vacuously true, and treats the false hypothesis as true, yielding a contradiction that by ex contradictione quodlibet implies and does not imply everything. So, such a conditional statement is true and false simultaneously.

Third Proof.Premises:s is a statement. Under the assumption that "s and it is not true that s," s is true. It is not true that: under the assumption that "s and it is not true that s," s is true.Conclusion: The absolute Russell set exists.

Statements (Reasons)1. s is a statement. (Premise)2. Under the assumption that "s and it is not true that s," s is true. (Premise)3. It is not true that: under the assumption that "s and it is not true that s," s is true. (Premise)4. Some contradiction exists. ((3) is the negation of (2))5. The absolute Russell set exists. (Ex contradictione quodlibet)This concludes the third proof.

Premise (1) is a prescribed description of s. Each of premises (2) and (3) is evident by itself.

Lemma. If a statement is false, then it materially implies some contradiction exists.

Note: If there is a statement "A", and if it is known that this statement "A" can be proven false, this produces the implication of a far broader statement than the mere denial of "A".

The capacity to indicate the falsehood of a statement implies a lot more than the truth or falsehood of that statement. An implication is a power. All sets are origins of powers. x is the power to be identified as x, so x is always a set. Sets underlie indicators, integers and all positives.

Every proof by contradiction of the nonexistence of the absolute Russell set includes a temporary assumption that implies, regardless of whether the assumption is in effect, the absolute Russell set exists.

Fourth Proof. Consider a proof by contradiction of the nonexistence of the absolute Russell set. Name the proof p. As my post at viewtopic.php?p=2699066#p2699066, which I have now cited six times in this thread, allows me to infer, under p's assumption that the absolute Russell set exists, the absolute Russell set is an element of the set. The linked post also allows me to infer that it is not true that under p's assumption that the absolute Russell set exists, the absolute Russell set is an element of the set. So, there is a contradiction. Thus, through ex contradictione quodlibet, the absolute Russell set exists. That concludes the fourth proof.

Every proof by contradiction that the absolute Russell set does not exist thus becomes a part of a proof that the set does exist.